Acceleration & the non-uniqueness of the unit normal vector to a curve at a point

Question

Knowing the equation of a curve in three-dimensional space in the following parametric form $$x=x(u), y=y(u), z=z(u),$$ we can determine the tangent vector $\hat{T}$ at a point $(x,y,z)$ using the formula $$\hat{T}(u)=\frac{\vec{r}^\prime(u)}{\sqrt{x^{\prime2}(u)+y^{\prime2}(u)+z^{\prime2}(u)}}.$$ Having obtained $\hat{T}(u)$, a fair task is to obtain a unit normal vector $\hat{N}(u)$ at that point. But the unit vector $\hat{N}(u)$ does not seem to be unique. The direction of $\hat{N}(u)$ can have any direction on the plane transverse to $\hat{T}(u)$. This troubles me! Let me say why.

From kinematics, the instantaneous acceleration of a particle moving along a trajectory $\vec{r}(t)$ are respectively given by $$\vec{a}(t)=\dot{v}(t)\hat{T}(t)+\kappa v^2(t)\hat{N}(t)$$ which tells that acceleration will not be unique unless $\hat{N}(t)$ is uniquely determined.

• But how is this possible that the acceleration to a given trajectory is not unique? What am I missing?

Answer 1

$\hat N(t)$ refers to the principal normal; the normal in the direction $\dfrac{d\hat T}{dt}$. More precisely, $$ \hat N(t)=\frac{\frac{d\hat T}{dt}}{\left\lVert\frac{d\hat T}{dt}\right\rVert} $$

Answer 2

Whereas it is certainly true that there are an infinite number of unit normal vectors $\hat n(t)$ to the curve

$\vec r(t) = (x(t), y(t), z(t)) \tag 1$

in the sense that

$\hat n(t) \cdot \hat T(t) = 0, \tag 2$

all of which lie on the unit circle in the plane perpendicular to $\hat T(t)$, there is in general only one unit vector $\hat N(t)$ directly related to the acceleration vector $\vec a(t)$ of the curve $\vec r(t)$, and that is defined by the first Frenet-Serret equation

$\dfrac{d\hat T(t)}{ds} = \kappa(t) \hat N(t), \tag 3$

where $\kappa(t)$ is the curvature of $\vec r(t)$, also defined by this equation; here the derivative of $\hat T(t)$ is taken with respect to arc-length $s$, though $\hat T(t)$ is expressed in terms of the given curve parameter $t$; to express this equation in terms of the $t$ derivatives of $\hat T(t)$ the factor $ds/dt$ is introduced into (3), and then the chain rule yields

$\dfrac{1}{\dfrac{ds}{dt}} \dfrac{d\hat T(t)}{dt} = \dfrac{dt}{ds} \dfrac{d\hat T(t)}{dt} = \kappa(t) \hat N(t). \tag 4$

The arc-length parameter $s$ is also a function of $t$; since it metrics distance along the curve $\vec r(t)$, in the usual manner it is given by the integral of the speed $v(t)$ with respect to $t$, where

$v(t) = \left \vert \dfrac{d \vec r(t)}{dt} \right \vert; \tag 5$

thus, the arc-length traversed 'twixt times $t_0$ and $t$ is

$s(t) = \displaystyle \int_{t_0}^t v(w) \; dw = \int_{t_0}^t \left \vert \dfrac{d \vec r(w)}{dw} \right \vert \; dw, \tag 6$

which of course immedtiately implies

$\dfrac{ds(t)}{dt} = v(t) = \left \vert \dfrac{d \vec r(t)}{dt} \right \vert, \tag 7$

in light of which (4) becomes

$\dfrac{1}{v(t)} \dfrac{d\hat T(t)}{dt} = \kappa(t) \hat N(t), \tag 8$

that is

$\dfrac{d\hat T(t)}{dt} = v(t) \kappa(t) \hat N(t). \tag 9$

Now the velocity vector

$\vec v(t) = \dfrac{d \vec r(t)}{dt} \tag{10}$

is tangent to the curve $\vec r(t)$, and since $T(t)$ is the unit tangent vector to this curve, we may write

$\vec v(t) = \dfrac{\vec dr(t)}{dt} = v(t)T(t) \tag{11}$

in accord with (5); it follows that

$\vec a(t) = \dfrac{d \vec v(t)}{dt} = \dfrac{d}{dt}(v(t)T(t))$ $= \dfrac{dv(t)}{dt}T(t) + v(t) \dfrac{dT(t)}{dt} = \dfrac{dv(t)}{dt}T(t) + v^2(t) \kappa(t) \hat N(t), \tag{12}$

where we have substituted in (9) to obtain the right-most equality; this is the formula for acceleration given in the text of the question; we observe it holds for that unique unit normal $\vec N(t)$ given in the Frenet-Serret formula (3).

Finally, a word or two about $\kappa(t)$; it is generally assumed in writing (3) that

$\kappa (t) \ne 0; \tag{13}$

otherwise, (3) cannot be determine/define $\hat N(t)$; thus the assumption (3) is generally made regarding curves $\vec r(t)$ to which the Frenet-Serret system is applied. The acceleration of the curve $\vec r(t)$ as given by (12) is thus unique as long as (13) binds. Of course, we may always define $\vec a(t)$ via

$\vec a(t) = \dfrac{d\vec v(t)}{dt} = \dfrac{d^2 \vec r(t)}{dt^2}, \tag{14}$

whether or not (13) is assumed.